Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Values vary for different materials, but for most metals, Poisson’s ratio has a value
between 1/4 and 1/3. Because the volume of material must remain constant, the largest
possible value for Poisson’s ratio is 0.5. Values approaching this upper limit are found only
for materials such as rubber.
57
POISSON’S RATIO
Relationship Between E, G, and ν
Poisson’s ratio is related to the elastic modulus E and the shear modulus G by the formula
G =
E
(3.7)
2(1 + ν)
The Poisson effect exhibited by
materials causes no additional
stresses in the lateral direction
unless the transverse deformation
is inhibited or prevented in some
manner.
ExAmpLE 3.1
A tension test was conducted on a 1.975 in. wide by 0.375 in. thick specimen of a nylon
plastic. A 4.000 in. gage length was marked on the specimen before application of the
load. In the elastic portion of the stress–strain curve at an applied load of P = 6,000 lb, the
elongation in the gage length was measured as 0.023 in. and the contraction in the width
of the bar was measured as 0.004 in. Determine
(a) the elastic modulus E.
(b) Poisson’s ratio ν.
(c) the shear modulus G.
1.975 in.
4.000 in.
P
Plan the Solution
(a) From the load and the initial measured dimensions of the bar, the normal stress can
be computed. The normal strain in the longitudinal (i.e., axial) direction, ε long , can
be computed from the elongation in the gage length and the initial gage length. With
these two quantities, the elastic modulus E can be calculated from Equation (3.4).
(b) From the contraction in the width and the initial width of the bar, the strain in the
lateral (i.e., transverse) direction, ε lat , can be computed. Poisson’s ratio can then be
found from Equation (3.6).
(c) The shear modulus can be calculated from Equation (3.7).
P
SOLUTION
(a) The normal stress in the plastic specimen is
The longitudinal strain is
Therefore, the elastic modulus is
6,000 lb
σ = = 8,101.27 psi
(1.975 in.)(0.375 in.)
0.023 in.
ε long = = 0.005750 in./in.
4.000 in.
E
σ 8,101.27 psi
= = = 1, 408,916 psi = 1,409,000 psi Ans.
ε 0.005750 in./in.